Executive summary: If we can bound the derivatives of some multivariate function with respect a continuous variable, do we automatically get bounds on the behavior of the maximum of that function with respect to other variables?
Here's the setup. Let $K$ be a compact topological space and let $f:\mathbb R\times K\to\mathbb R$ be a continuous function. Suppose the partial derivative $\frac{\partial}{\partial t}f(t, x)$ exists everywhere and, in fact, defines a continuous function $f':R\times K\to\mathbb R$. Define a function $Mf:\mathbb R\to\mathbb R$ by $Mf(t)=\max_{x\in K} f(t, x)$. Then $Mf$ is continuous, I'm pretty sure, but not necessarily differentiable. Nonetheless, the derivatives $f'$ evaluated at the witnesses of $Mf$ should bound the variation of $Mf$. Namely, if we focus our attention on $t$ around $0$ for simplicity:
- Let $MK(0)\subset K$ be the set of witnesses for $Mf(0)$. That is, $MK(0)$ is the set of $x\in K$ such that $f(0, x) = Mf(0) = \max_{y\in K}f(0, y)$. Or, if this notation is available, $MK(0)=\mathrm{Argmax}_{x\in K}f(0, x)$.
- Since $MK(0)$ is itself a compact set, we may define $Mf^+(0)=\max_{x\in MK(0)}f'(0, x)$ and $Mf^-(0)=\min_{x\in MK(0)}f'(0, x)$.
Then we should have the following conclusion:
- For all $\epsilon>0$, there exists $\delta > 0$ such that for all $t\in(-\delta,\delta)$, $t\neq 0$, the inequalities hold:
$$ Mf^-(0)-\epsilon < \frac{MF(t)-MF(0)}{t} < Mf^+(0)+\epsilon $$
This feels like it ought to be an easy, well-known theorem, but I lack the terminology to find it.
If it helps, we can assume that $f'$ has a uniform bound, or that it has higher-order partial derivatives. Or we can restrict the domain of $t$ from $\mathbb R$ to $[-1,1]$. We can also assume that $K$ obeys separation axioms, but not that it's connected. For example, I want this result to hold if $K$ is a Cantor set.
Note that my question asks for something like a generalization of Interchanging the order of maximization and diffrentation.